From the hint, we solve the equation $\displaystyle \vec{y}=A(B\vec{x})$ first for $\displaystyle B(\vec{x})$ and then for $\displaystyle \vec{x}$. So we have

$\displaystyle B\vec{x}=A^{-1}\vec{y}$

so...

$\displaystyle \vec{x}=B^{-1}A^{-1}\vec{y}$

I'm really not sure where to go from here though.

Mar 1st 2011, 06:30 PM

Ackbeet

Well, what does the fact that both $\displaystyle A$ and $\displaystyle B$ are invertible (so you can write their inverses) say about the last equation you wrote down? Is the transformation invertible?

Mar 1st 2011, 07:01 PM

centenial

Ok, I think I understand.

So it is invertible and its inverse is simply:

$\displaystyle \vec{x} = B^{-1}A^{-1}\vec{y}$

Is that correct?

Mar 1st 2011, 07:10 PM

Ackbeet

I suppose your answer is in keeping with the question's nomenclature in the OP. I would say that $\displaystyle AB$ is the linear transformation, and that